HMWK 5 Ch 17: P 6, 11, 30, 31, 34, 42, 50, 56, 58, 60 Ch 18: P 7, 16, 22, 27, 28, 30, 51, 52, 59, 61 Ch. 17 P17.6. Prepare: The laser beam is an electromagnetic wave that travels with the speed of light. We will first find the speed of light in the unknown liquid and then use Equation 17.1 to find its index of refraction. Finally, we will use Equation 17.2 to find the wavelength of light in the unknown liquid. The speed of light in the liquid is v liquid 30 10 2 m 1.38 10 9 s 2.174 108 m/s The liquid s index of refraction is n to be reported as 1.4 to two significant figures. Thus the wavelength of the laser beam in the liquid is c 3.0 108 v liquid 2.174 10 8 1.38 λ liquid λ vac n 633 nm 460 nm 1.38 The wavelength falls in the visible region. P17.11. Prepare: Two closely spaced slits produce a double-slit interference pattern. The interference pattern looks like the photograph of Figure 17.9. Note that there are 11 gaps between a span of 12 fringes or a distance of 52 mm. The formula for the fringe spacing is y λl d 52 10 11 3 m (633 10 9 m)(3.0 m) d 0.40 mm d This is a reasonable distance between the slits, ensuringd/l 1.34 10 4 << 1. P17.30. Prepare: A narrow slit produces a single-slit diffraction pattern. The intensity pattern for single-slit diffraction will look like Figure 17.27 with minima given by Equation 17.24. The width of the central maximum is given by Equation 17.25. The width of the central maximum for a slit of widtha 200λ is w 2λL a 2(500 10 9 m)(2.0 m) 0.0040 m 4.0 mm 0.0005 m
P17.31. Prepare: A narrow slit produces a single-slit diffraction pattern. The intensity pattern for single-slit diffraction will look like Figure 17.27. Angleθ 0.70 0.0122 rad is a small angle (<< 1 rad). Thus we use Equation 17.23 to find the wavelength of light. The angles of the minima of intensity are θ p p λ a p 1, 2, 3,... λ aθ p p (0.10 10 3 m)(0.0122 rad) 2 610 nm The wavelength is in the visible region, so its value is reasonable. P17.34. Prepare: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. The intensity pattern will look like Figure 17.31. The diameter D can be found from Equation 17.27. From Equation 17.27, the diameter of the circular aperture is D 2.44λL w 2.44(633 10 9 m)(4.0 m) 2.5 10 2 m 0.25 mm The diameter obtained above is typical for observing diffraction. P17.42. Prepare: For a double slit, the location of the mth interference maxima is determined by y m mλl /d. The wavelength of light in a medium with an index of refraction of n is related to the wavelength in air byλ n λ air /n. (a) The spacing between the second and third bright fringes under water is ( y) water (y 3 ) water (y 2 ) water (3 2)λ water L /d (4.5 10 7 m)(0.9 m)/0.34 10 3 m 1.35 mm (b) When the water is drained, the wavelength will increase by a factor of n (λ air nλ water ). When this expression is inserted into the expression for the location of the interference fringe maxima we have (y m ) air mλ air L/d m(nλ water )L/d. Using this expression the spacing between the second and third bright fringes in air is ( y) air nλ water L/d n( y) water (1.33)(1.35 mm) 1.80 mm Once you get into the lab and use a single slit, you will see that this is a reasonable fringe spacing.
P17.50. Prepare: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. The angle of diffraction is determined by the constructive-interference condition d sinθ m mλ, where m 0, 1, 2, 3,... If blue light (the shortest wavelengths) is diffracted at angle θ, then red light (the longest wavelengths) is diffracted at angle θ + 30. (a) In the first order, the equations for the blue and red wavelengths are sinθ λ b d Combining the two equations we get for the red wavelength, d sin(θ + 30 ) λ r λ r d (sinθ cos30 + cosθ sin 30 ) d(0.86 60sinθ + 0.50 cosθ ) d λ b d 0.8660 + d(0.50) 1 λ 2 b d 2 (0.50)d 1 λ 2 b λ d 2 r 0.8660 λ b (0.50) 2 (d 2 2 λ b ) (λ r 0.8660λ b ) 2 d λ r 0.866 0λ b 0.50 2 + λ b 2 Usingλ b 400 10 9 m andλ r 700 10 9 m, we getd 8.125 10 7 m. This value of d corresponds to 1mm d 1.0 10 3 m 1230 lines/mm 8.125 10 7 m (b) Using the value of d from part (a) andλ 589 10 9 m, we can calculate the angle of diffraction as follows: d sinθ 1 (1)λ (8.125 10 7 m)sinθ 1 589 10 9 m θ 1 46.5 1230 lines/mm for the diffraction grating is reasonable. P17.56. Prepare: The relationship between the diffraction grating spacing(d), the angle at which a particular order of constructive interference occurs(θ m ), the wavelength of the light, and the order of the constructive interference(m) is d sinθ m mλ and N 1/d. The first order diffraction angle for green light is θ 1 sin 1 (λ/d) sin 1 (5.5 10 7 m/2.0 10 6 m) sin 1 (0.275) 0.278 radian 16 This is a reasonable angle for the first order maximum.
P17.58. Prepare: A diffraction grating produces an interference pattern like the diagram of Figure 17.12. The angle of diffraction is determined by the constructive-interference condition, Equation 17.12,d sinθ m mλ, where m 0, 1, 2, 3,... Furthermore, according to Equation 17.13, the distance y m from the center to the mth maximum is y m L tanθ m. A key statement is that the lines are seen on the screen. This means that the light is visible light, in the range 400 nm 700 nm. We can determine where the entire visible spectrum falls on the screen for different values of m. We do this by finding the angles θ m at which 400 nm light and 700 nm light are diffracted. We then use y m L tanθ m to find their positions on the screen, which is at a distance L 75 cm. (a) The slit spacing isd 1 mm/1200 8.333 10 7 m. For m 1, For m 2, λ 400 nm:θ 1 sin 1 (400 nm/d) 28.7 y 1 75 cm tan28.7 41cm λ 700 nm: θ 1 sin 1 (700 nm/d) 57.1 y 1 75 cm tan57.1 116 cm λ 400 nm:θ 1 sin 1 (2 400 nm/d) 73.8 y 2 75 cm tan73.8 257 cm For the 700 nm wavelength at m 2,θ 2 sin 1 (2 700 nm/d) sin 1 (1.68) is not defined, so y 2. We see that visible light diffracted at m 1 will fall in the range 41 cm y 116 and that visible light diffracted at m 2 will fall in the range y 257 cm. These ranges do not overlap, so we can conclude with certainty that the observed diffraction lines are all m 1. (b) To determine the wavelengths, we first find the diffraction angle from the observed position by using y y θ tan 1 tan 1 L 75 cm This angle is then used in the diffraction grating equation for the wavelength with m 1,λ dsinθ 1. Line θ λ 56.2 cm 36.85 500 nm 65.9 cm 41.30 550 nm 93.5 cm 51.27 650 nm
P17.60. Prepare: Reflection is maximized for constructive interference of the two reflected waves, but minimized for destructive interference. The first reflection occurs at the front surface of the oil film. Here, the index of refraction increases from that of air (n 1) to that of the film (n 1.42), so there will be a reflective phase change. The light then reflects from the rear surface of the film. The index of refraction again increases from 1.42 to 1.50 for the glass, and there is a reflective phase change. With two phase changes, Tactics Box 17.1 tells us that we should use Equation 17.15 for constructive interference and Equation 17.16 for destructive interference. (a) Constructive interference of the reflected waves occurs for wavelengths given by Equation 17.15: λ m 2nt m 2(1.42)(500 nm) m (1420 nm) m Thus, λ 1 1420 nm, λ 2 1 2 (1420 nm) 710 nm, λ 3 473 nm, λ 4 355 nm,... Only the wavelength of 473 nm is in the visible range.(b) For destructive interference of the reflected waves, λ 2nt m+ 1 2 2(1.42)(500 nm) m + 1 2 1420 nm m+ 1 2 Thus, λ 0 2 1420 nm 2840 nm, λ 1 2 3 (1420 nm) 947 nm), λ 2 568 nm, λ 3 406 nm,... The wavelengths of 406 nm and 568 nm are in the visible range. (c) Beyond the limits 430 nm and 690 nm the eye s sensitivity drops to about 1 percent of its maximum value. The reflected light is enhanced in blue (473 nm). The transmitted light at mostly 568 nm will be yellowish green. Ch. 18 P18.7. Prepare: We ll use the law of reflection and see where rays of light coming from object O and reflecting in the mirror can go. The figure shows rays coming from O and hitting each end of the mirror. It is clear that object O can be seen in the mirror from any location between the outer rays, which includes locations B and C. However, at location A object O can t be seen in the mirror because there is no way for a ray to obey the law of reflection from O and get to A. If the mirror extended further to the left, then A could see O. A similar argument can be made that object O can t be seen in the mirror from position D. In summary, the object s image is visible from B and C. tells us. Your intuition in looking at the original diagram confirms what the ray tracing
P18.16. Prepare: Use the ray model of light. The figure incorporates the first four steps of Tactics Box 18.1. θ dia. Using Snell s law at the water-diamond boundary, n dia sinθ dia n water sinθ water sinθ water n dia sinθ dia 2.41 sin30 0.9060 θ water 65 n water 1.33 Because n dia is much larger than n water, we expectedθ water to be much larger than P18.22. Prepare: We know that violet light is refracted more than red (this is reflected in the given values for n), so we draw a diagram with the violet ray in the glass closer to the normal then the red ray. But they emerge together at the same point and in the same direction. We are givenn red 1.54, n violet 1.59, andθ air 22.5. We also known air 1.00. The strategy will be to solve Snell s law for θ violet andθ red in the glass and then find θ as the difference of the two angles. n air sinθ air n violet sinθ violet n red sinθ red Solve Snell s law forθ violet andθ red in the glass. n θ vi ol et sin 1 ai r sin θ air 1.00 sin 22.5 sin 1 1.59 13.93 n vi ole t n θ red sin 1 ai r sinθ a ir 1.00 sin 22.5 sin 1 1.54 1 4.39 n re d The difference between these angles (i.e., the angle between the two rays in the glass) is θ 14.39 13.93 0.462. The angle between the two rays is very small, but that is to be expected since the index of refraction isn t very different for the two rays.
P18.27. Prepare: Use ray tracing to locate the image. The figure below shows the ray-tracing diagram using the steps of Tactics Box 18.2. You can see from the diagram that the image is in the plane where the three special rays converge. The image is located at s 15 cm to the right of the converging lens, and is inverted and real. Ray tracing must be done to scale to obtain useful answers. P18.28. Prepare: Use ray tracing to locate the image. The figure shows the ray-tracing diagram using the steps of Tactics Box 18.2. You can see that the rays after refraction do not converge at a point on the refraction side of the lens. On the other hand, the three special rays, when extrapolated backward toward the incidence side of the lens, meet at P which is 15 cm from the lens. That is, s 15 cm. The image is upright and virtual. Ray tracing must be done to scale to obtain useful answers.
P18.30. Prepare: Use ray-tracing to locate the image. The figure shows the ray tracing diagram using the steps of Tactics Box 18.2. The three rays after refraction do not converge at a point, but they appear to come from P. P is 6 cm from the diverging lens, so s 6 cm. The image is upright and virtual. Draw ray tracings accurately to scale. P18.51. Prepare: This is both a refraction (Snell s law) problem and a critical angle for total internal reflection problem. Label the normal lines on your sketch carefully. We first compute the critical angle inside the fiber optic core using Equation 18.3. The critical angle is measured from the normal, which is vertical in this drawing. θ c sin 1 n 2 sin 1 1.45 75 1.50 n 1 Now look at Figure P18.51 carefully. The 75 from the normal on the side with the cladding means the angle for refraction from the air-glass interface (measured from the new normal at the left) is90 75 15. θ θ a sin 1 n g sinθ g n a sin 1 1.50 sin 15 23 1.00 The numbers all seem reasonable, but the diagram is clearly not drawn to scale as θ is depicted as much greater than23, but that s OK.
P18.52. Prepare: Use the ray model of light and the law of refraction. Assume that the laser beam is a ray of light. The laser beam enters the water 2.0 m from the edge, undergoes refraction, and illuminates the goggles. The ray of light from the goggles then retraces its path and enters your eyes. We will use Snell s law at the airwater boundary and the geometry of the following diagram. From the geometry of the diagram, tanφ 1.0 m 2.0 m φ tan 1 (0.50) 26.57 θ air 90 26.57 63.43 Snell s law at the air-water boundary isn air sinθ air n water sinθ water. Using the above result, (1.0)sin 63.43 1.33 sinθ water θ water sin 1 sin 63.43 1.33 42.26 Taking advantage of the geometry in the diagram again, x 3.0 m tanθ water x (3.0 m) tan 42.26 2.73 m The distance of the goggles from the edge of the pool is 2.73 m + 2.0 m 4.73 m or 4.7 to two significant figures. The given distances and dimension measurements for the pool certainly indicate that the result obtained above is reasonable. P18.59. Prepare: We will use the ray model of light and apply Snell s law. Please refer to Figure P18.59. (a) Using Snell s law at the air-glass boundary, with φ being the angle of refraction inside the prism, n air sinβ n sinφ sinβ n sinφ Considering the triangle made by the apex angle and the refracted ray, Thus (b) Using the above expression, we obtain n sinβ sin 52.2 sin ( 1 2 α) sin 30 1.58 (90 φ) + (90 φ) + α 180 φ 1 2 α sinβ n sin 1 2 α β sin 1 n sin 1 2 α
P18.61. Prepare: Represent the aquarium s wall as a point source, and use the ray model of light. Paraxial rays from the outer edge (O) are refracted into the water and then enter into the fish s eye. When extended into the wall, these rays will appear to be coming from O rather from O. The point on the inside edge (I) of the wall will not change its apparent location. We are given that s O s I 4.00 mm and s O s I 3.50 mm. Using Equation 18.7, s O n water s n O s I n water s wall n I wall s O s I n water (s n O s I ) 3.50 mm 1.33 4.00 mm (4.00 mm) n wall n wa ll (1.33) 1.52 wall 3.5 0 mm The value obtained above is closer to that of glass, so it is reasonable.